Tour:Subgroup containment relation equals divisibility relation on generators

This article adapts material from the main article: subgroup containment relation in the group of integers equals divisibility relation on generators

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WHAT YOU NEED TO DO:
• Read and understand the main statement, as well as the corollaries.
• Convince yourself of the truth of the main statement, as well as of how the corollaries follow from it.

Statement

Let $\mathbb{Z}$ denote the Group of integers (?) under addition. Suppose $H,K$ are subgroups of $\mathbb{Z}$. Suppose $H = m \mathbb{Z}$ and $K = n \mathbb{Z}$. Then, $H \le K$ if and only if $n| m$, i.e., $m$ is a multiple of $n$.

Note that, because every nontrivial subgroup of the group of integers is cyclic on its smallest element, the subgroups $H$ and $K$ can be written in the form $m\mathbb{Z}, n\mathbb{Z}$ respectively.

Related facts

Corollaries

• Given two subgroups $m\mathbb{Z}$ and $n\mathbb{Z}$, their intersection is the subgroup generated by an element $l$ with the property that $m | l, n | l$, and if $m | k, n|k$, then $l|k$. Such an integer $l$ is termed a least common multiple of $m$ and $n$ (if we allow only nonnegative integers, then it is unique).
• Given two subgroups $m\mathbb{Z}$ and $n\mathbb{Z}$, their join is the subgroup generated by an element $d$ with the property that $d | m, d | n$, and if $c | m, c|n$, then $c | d$. Such an integer $d$ is termed a greatest common divisor of $m$ and $n$ (if we allow only nonnegative integers, then it is unique).
• The greatest common divisor of $m$ and $n$ can be written as $am + bn$ for some integers $a$ and $b$. That is because it is in the subgroup generated by $m\mathbb{Z}$ and $n\mathbb{Z}$.